使用Linq在列表中查找连续项 [英] Finding Consecutive Items in List using Linq
问题描述
说我有以下整数数组:
int[] numbers = { 1, 6, 4, 10, 9, 12, 15, 17, 8, 3, 20, 21, 2, 23, 25, 27, 5, 67,33, 13, 8, 12, 41, 5 };
我如何编写一个Linq查询来查找3个大于10的连续元素?另外,如果我可以指定我要说的是此类元素的第一,第二,第三等组,那就太好了.
How could I write a Linq query that finds 3 consecutive elements that are, say, greater than 10? Also, it would be nice if I could specify I want say the first, second, third etc. group of such elements.
例如,Linq查询应该能够识别: 12,15,17作为第一组连续元素 第二组23,25,27 第三组67,33,13
For example, the Linq query should be able to identify: 12,15,17 as the first group of consecutive elements 23,25,27 as the second group 67,33,13 as the third group
如果我指定我希望第二组包含3个连续元素,则查询应返回第二组.
The query should return to me the 2nd group if I specify I want the 2nd group of 3 consecutive elements.
谢谢.
推荐答案
更新:虽然从技术上讲不是Patrick指出的"linq查询",但此解决方案可重用,灵活且易于使用.通用的.
UPDATE: While not technically a "linq query" as Patrick points out in the comments, this solution is reusable, flexible, and generic.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace ConsoleApplication32
{
class Program
{
static void Main(string[] args)
{
int[] numbers = { 1, 6, 4, 10, 9, 12, 15, 17, 8, 3, 20, 21, 2, 23, 25, 27, 5, 67,33, 13, 8, 12, 41, 5 };
var consecutiveGroups = numbers.FindConsecutiveGroups((x) => x > 10, 3);
foreach (var group in consecutiveGroups)
{
Console.WriteLine(String.Join(",", group));
}
}
}
public static class Extensions
{
public static IEnumerable<IEnumerable<T>> FindConsecutiveGroups<T>(this IEnumerable<T> sequence, Predicate<T> predicate, int count)
{
IEnumerable<T> current = sequence;
while (current.Count() > count)
{
IEnumerable<T> window = current.Take(count);
if (window.Where(x => predicate(x)).Count() >= count)
yield return window;
current = current.Skip(1);
}
}
}
}
输出:
12,15,17
23,25,27
67,33,13
要获得第二组,请更改:
To get the 2nd group, change:
var consecutiveGroups = numbers.FindConsecutiveGroups((x) => x > 10, 3);
收件人:
var consecutiveGroups = numbers.FindConsecutiveGroups((x) => x > 10, 3).Skip(1).Take(1);
更新2 在我们的生产用途中对此进行了调整之后,随着numbers数组中的项数越来越大,下面的实现要快得多.
UPDATE 2 After tweaking this in our production use, the following implementation is far faster as the count of items in the numbers array grows larger.
public static IEnumerable<IEnumerable<T>> FindConsecutiveGroups<T>(this IEnumerable<T> sequence, Predicate<T> predicate, int sequenceSize)
{
IEnumerable<T> window = Enumerable.Empty<T>();
int count = 0;
foreach (var item in sequence)
{
if (predicate(item))
{
window = window.Concat(Enumerable.Repeat(item, 1));
count++;
if (count == sequenceSize)
{
yield return window;
window = window.Skip(1);
count--;
}
}
else
{
count = 0;
window = Enumerable.Empty<T>();
}
}
}
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